Question

Let \(f:\mathbb R \to \mathbb R\) be a function such that for any irrational number \(r\), and any real number \(x\) we have \(f(x)=f(x+r)\). Show that \(f\) is a constant function.

Answer 1

This seems to be the case just in general for any function that satisfies

\[f(a)=f(a+b)\]

then f(x) would have to be constant.

Answer 2

but what about proving for real numbers?

Answer 3

@jtvatsim sounded like he knew more about this than I did, I was just making a passing comment. I don't really study this sort of math so I have no idea about density of numbers or whatever.

Answer 4

@empty "sounded" is the key word. :) I'm trying to make this more rigorous. lol

Answer 5

I think I got it, thanks everyone

Answer 6

What was the key?

Answer 7

the question really only asks me to prove that whatever I put into the function, I will get out \(r\), and so I can do that and spin off something similar to what empty said.

Answer 8

It is sufficient to show that \[f(x+\epsilon) - f(x) = 0\]

By the given definition the left hand side is same as
\[f(x+r+\epsilon) - f(x) \]
for any irrational \(r\).
let \(r,\epsilon\) be positive for simplicity

since anything added to iraational gives an irrational, the first term evaluates to \(f(x)\)

Answer 9

Convincing to me... QED.

Answer 10

But wait... This depends on the fact that irrational + anything is also irrational... is this necessarily true?

Answer 11

yeah below is bugging me a bit \[\sqrt{2}+ (-\sqrt{2}) \]

Answer 12

Indeed... that is troubling. :/

Answer 13

thats not really a problem because

\[f(x+\sqrt{2}+(-\sqrt{2})) = f(x+0) = f(x)\]

:P

Answer 14

But we can make it a problem easily... :P lol

Answer 15

the problem comes only when \(\epsilon\) and \(r\) are additive inverses

Answer 16

Consider \[2.dec(\sqrt{2}) - 0.dec(\sqrt{2}) = 2\]
where \[dec(\sqrt{2})\] uses the digits from \[\sqrt{2}\].

Answer 17

true! i take my statement back :)

Answer 18

I think we can get around that issue by simply fixing the irrational, lets say, \(r=\sqrt{2}\)

we won't be losing anything.

Answer 19

Let \(\epsilon\) be some positive real number.
It is sufficient to show that \[f(x+\epsilon) - f(x) = 0\]

By the given definition the left hand side is same as
\[f(x+\color{red}{\sqrt{2}}+\epsilon) - f(x) \]

since \(\sqrt{2}+\epsilon\) is irrational for all positive \(\epsilon\), the first term evaluates to \(f(x)\)

Answer 20

nice work

Answer 21

I'm going to have to keep thinking through this one. Your new approach is clever, but I feel there is a way to force sqrt(2) + irrational to still be rational. It all depends on if the sequence of digits (10-d) where d is the decimal digit of sqrt(2) is repeating or nonrepeating... however, I don't think I have the firepower at this point to deal with that. Too much fun! :)

Answer 22

Ahh you want to make
\[\epsilon = \text{10'scompliment}(\sqrt{2}-1)\]
very interesting xD

Answer 23

Basically, it's hard to formalize.... However, I have this suspicion that even this would work in our favor since x can be any real number, we might be able to pair up our x's and epsilons in order to cover all real number possibilities.... I may be overthinking this xP

Answer 24

Are we assuming here that f is continuous? That would be helpful I imagine to simplify all this crazy infinitesimal stuff...

Answer 25

I thought earlier you're saying something like below
\[\begin{align} \sqrt{2}+\epsilon &= \sqrt{2} + \text{9'scompliment}(\sqrt{2}-1)\\~\\
&=1.413213\ldots + 0.586786\ldots \\~\\
&=1.999999\ldots\\~\\
&=2
\end{align}\]

Answer 26

Yes, precisely. Sorry, my train of thought wasn't necessarily connected. :)

Answer 27

I lack that topological property in general...

Answer 28

lol this is indeed very interesting, never thought of this trick before

this observation essentially implies that sum of two positive irrational numbers need not be irrational ?

Answer 29

Indeed, which is what I was afraid of. It is a very disturbing thought. :)

Answer 30

nvm i wasn't even thinking, see :
\[(5-\sqrt{2})+( 5+\sqrt{2}) = 10\]

Answer 31

Oh duh... I can think complicated, just can't come up with simple counter examples. :P lol

Answer 32

There is an interesting proof out there that if x and y are irrational, then precisely one of x + y or x - y is also irrational.

Answer 33

that invalidates the proof, so we do need to consider two cases

Answer 34

We definitely know that the function evaluated at all the irrationals is the same.

Let x = 0. Then f(0) = f(r) for all irrational r.

Answer 35

that proves the function is constant for all irrational numbers, r

Answer 36

Let \(\epsilon\) be some positive real number.
It is sufficient to show that \[f(x+\epsilon) - f(x) = 0\]

By the given definition the left hand side is same as
\[f(x+\color{red}{\sqrt{2}}+\epsilon) - f(x) \]

If \(\epsilon\) is rational, then \(\sqrt{2}+\epsilon\) is irrational, so the first term evaluates to \(f(x)\)

If \(\epsilon\) is irrational, then \(f(x+\color{red}{\sqrt{2}}+\epsilon) = f(x+\color{red}{\sqrt{2}}) = f(x)\) . \(\blacksquare\)

Answer 37

Ooh ooh! How about this!

Let x = p - r where p is rational. Now we DO KNOW that rational + irrational is irrational. Thus, x is an irrational.

Then,
f(x) = f(p - r) = f(p) for all rationals p.

But all irrationals are equal to f(0).

Hence,

f(r) = f(p) = f(0) as desired.

Answer 38

That looks really neat !

Answer 39

Finally, we have something that works.... that was a real pain. Don't know why it was so elusive...

Answer 40

Haha idk from where i got the idea that irrational+irrational = irrational

Answer 41

Well, I thought it was true at first too! I guess it seems so obvious that "really weird number + really weird number = really weird number"... but, not so, math thwarts our intuition once again! :)

Answer 42

Cool sounds like you guys are ready to solve one of these open questions: https://en.wikipedia.org/wiki/Irrational_number#Open_questions

Answer 43

I'm not going to even look... I will not be able to sleep tonight if I do.... :P But, I'll check it out at some point tomorrow. :D

Answer 44

there is a reason for calling them "open" lmao

Answer 45

\(\sqrt{2}+(2-\sqrt{2})\in \mathbb{Q}\)

Answer 46

seriously... props to everyone who stuck around! :)

Answer 47

Hey if there's anything I learned from set theory it's that something can be both open and closed at the same time @ganeshie8 :P

Answer 48

Actually it has nothing to do with set theory. Topology my man :)

Answer 49

What weird kind of definition are you using for topological space?

Answer 50

The most general one... Set theory, in general, has nothing to do with open/closed. Topology by definition is the analytic notion of open and closed sets. Show me one theorem in set theory about open closed. Then google "topology".